3.88 \(\int \frac {(d-c^2 d x^2)^{5/2} (a+b \sin ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=277 \[ \frac {5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac {5}{2} c^4 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 c^3 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt {1-c^2 x^2}}-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{6 x^2 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}-\frac {7 b c^3 d^2 \log (x) \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}} \]

[Out]

5/3*c^2*d*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x-1/3*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^3+5/2*c^4*d^2*
x*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)-1/6*b*c*d^2*(-c^2*d*x^2+d)^(1/2)/x^2/(-c^2*x^2+1)^(1/2)-1/4*b*c^5*d^2
*x^2*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+5/4*c^3*d^2*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/(-c^2*x^2+
1)^(1/2)-7/3*b*c^3*d^2*ln(x)*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4695, 4647, 4641, 30, 14, 266, 43} \[ \frac {5}{2} c^4 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 c^3 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt {1-c^2 x^2}}+\frac {5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {b c^5 d^2 x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{6 x^2 \sqrt {1-c^2 x^2}}-\frac {7 b c^3 d^2 \log (x) \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x^4,x]

[Out]

-(b*c*d^2*Sqrt[d - c^2*d*x^2])/(6*x^2*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^2*Sqrt[d - c^2*d*x^2])/(4*Sqrt[1 - c^2
*x^2]) + (5*c^4*d^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/2 + (5*c^2*d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSi
n[c*x]))/(3*x) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(3*x^3) + (5*c^3*d^2*Sqrt[d - c^2*d*x^2]*(a + b*A
rcSin[c*x])^2)/(4*b*Sqrt[1 - c^2*x^2]) - (7*b*c^3*d^2*Sqrt[d - c^2*d*x^2]*Log[x])/(3*Sqrt[1 - c^2*x^2])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(a + b*ArcSin[c*x])^(n - 1), x],
x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4695

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n)/(f*(m + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^
(m + 2)*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/
(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1),
x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x^4} \, dx &=-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {1}{3} \left (5 c^2 d\right ) \int \frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x^2} \, dx+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {\left (1-c^2 x^2\right )^2}{x^3} \, dx}{3 \sqrt {1-c^2 x^2}}\\ &=\frac {5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\left (5 c^4 d^2\right ) \int \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\left (1-c^2 x\right )^2}{x^2} \, dx,x,x^2\right )}{6 \sqrt {1-c^2 x^2}}-\frac {\left (5 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {1-c^2 x^2}{x} \, dx}{3 \sqrt {1-c^2 x^2}}\\ &=\frac {5}{2} c^4 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (c^4+\frac {1}{x^2}-\frac {2 c^2}{x}\right ) \, dx,x,x^2\right )}{6 \sqrt {1-c^2 x^2}}-\frac {\left (5 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (\frac {1}{x}-c^2 x\right ) \, dx}{3 \sqrt {1-c^2 x^2}}+\frac {\left (5 c^4 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}-\frac {\left (5 b c^5 d^2 \sqrt {d-c^2 d x^2}\right ) \int x \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{6 x^2 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}+\frac {5}{2} c^4 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac {5 c^3 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt {1-c^2 x^2}}-\frac {7 b c^3 d^2 \sqrt {d-c^2 d x^2} \log (x)}{3 \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.65, size = 243, normalized size = 0.88 \[ \frac {1}{24} d^2 \left (\frac {\sqrt {d-c^2 d x^2} \left (4 a \sqrt {1-c^2 x^2} \left (3 c^4 x^4+14 c^2 x^2-2\right )-56 b c^3 x^3 \log (c x)+b \left (-6 c^5 x^5+3 c^3 x^3-4 c x\right )\right )}{x^3 \sqrt {1-c^2 x^2}}-60 a c^3 \sqrt {d} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )+\frac {4 b \left (3 c^4 x^4+14 c^2 x^2-2\right ) \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{x^3}+\frac {30 b c^3 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)^2}{\sqrt {1-c^2 x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x^4,x]

[Out]

(d^2*((4*b*Sqrt[d - c^2*d*x^2]*(-2 + 14*c^2*x^2 + 3*c^4*x^4)*ArcSin[c*x])/x^3 + (30*b*c^3*Sqrt[d - c^2*d*x^2]*
ArcSin[c*x]^2)/Sqrt[1 - c^2*x^2] - 60*a*c^3*Sqrt[d]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))]
+ (Sqrt[d - c^2*d*x^2]*(4*a*Sqrt[1 - c^2*x^2]*(-2 + 14*c^2*x^2 + 3*c^4*x^4) + b*(-4*c*x + 3*c^3*x^3 - 6*c^5*x
^5) - 56*b*c^3*x^3*Log[c*x]))/(x^3*Sqrt[1 - c^2*x^2])))/24

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fricas [F]  time = 3.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^4,x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))*sqr
t(-c^2*d*x^2 + d)/x^4, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.48, size = 1527, normalized size = 5.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^4,x)

[Out]

5/2*a*c^4*d^3/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+4/3*a*c^2/d/x*(-c^2*d*x^2+d)^(7/2)+5/
3*a*c^4*d*x*(-c^2*d*x^2+d)^(3/2)-5/4*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)^2*d^2
*c^3-5/2*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)/(c^2*x^2-1)*c^3*(-c^2*x^2+1)^(1/2)-1/2*b*(-d*(
c^2*x^2-1))^(1/2)*d^2*c^4/(c^2*x^2-1)*arcsin(c*x)*x+1/3*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)
/x^3/(c^2*x^2-1)*arcsin(c*x)+1/2*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^6/(c^2*x^2-1)*arcsin(c*x)*x^3+7/3*b*(-d*(c^2*x
^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^2-1)*d^2*c^3+1/4*b*(-d*(c^2*x^2-1))^
(1/2)*d^2*c^5/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-1/3*a/d/x^3*(-c^2*d*x^2+d)^(7/2)+4/3*a*c^4*x*(-c^2*d*x^2+d)^(
5/2)+7/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c
^3-49/6*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^3/(c^2*x^2-1)*(-c^2*x^2+1)*c^6+7/6*I*b*(-d*
(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x/(c^2*x^2-1)*(-c^2*x^2+1)*c^4-35*I*b*(-d*(c^2*x^2-1))^(1/2)*
d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^5+147*I*b*(-d*(c^2*x^2-1))^(1/2
)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^7-1/8*b*(-d*(c^2*x^2-1))^(1/2
)*d^2*c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+21/2*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^2/(c^2*
x^2-1)*(-c^2*x^2+1)^(1/2)*c^5+190/3*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x/(c^2*x^2-1)*arcsi
n(c*x)*c^4-23/3*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)/x/(c^2*x^2-1)*arcsin(c*x)*c^2+1/6*b*(-d
*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)/x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c-14*I*b*(-c^2*x^2+1)^(1/
2)*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)*d^2*c^3/(3*c^2*x^2-3)-49/6*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15
*c^2*x^2+1)*x^5/(c^2*x^2-1)*c^8+28/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^3/(c^2*x^2-1)*
c^6-7/6*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x/(c^2*x^2-1)*c^4+147*b*(-d*(c^2*x^2-1))^(1/2
)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^5/(c^2*x^2-1)*arcsin(c*x)*c^8-203*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-1
5*c^2*x^2+1)*x^3/(c^2*x^2-1)*arcsin(c*x)*c^6+5/2*a*c^4*d^2*x*(-c^2*d*x^2+d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \sqrt {d} \int \frac {{\left (c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x^{4}}\,{d x} + \frac {1}{6} \, {\left (10 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d x + 15 \, \sqrt {-c^{2} d x^{2} + d} c^{4} d^{2} x + 15 \, c^{3} d^{\frac {5}{2}} \arcsin \left (c x\right ) + \frac {8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} c^{2}}{x} - \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}}}{d x^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^4,x, algorithm="maxima")

[Out]

b*sqrt(d)*integrate((c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1
)*sqrt(-c*x + 1))/x^4, x) + 1/6*(10*(-c^2*d*x^2 + d)^(3/2)*c^4*d*x + 15*sqrt(-c^2*d*x^2 + d)*c^4*d^2*x + 15*c^
3*d^(5/2)*arcsin(c*x) + 8*(-c^2*d*x^2 + d)^(5/2)*c^2/x - 2*(-c^2*d*x^2 + d)^(7/2)/(d*x^3))*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(5/2))/x^4,x)

[Out]

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(5/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*d*x**2+d)**(5/2)*(a+b*asin(c*x))/x**4,x)

[Out]

Integral((-d*(c*x - 1)*(c*x + 1))**(5/2)*(a + b*asin(c*x))/x**4, x)

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